Friction (braking)

Question

Car (1000 kg) brakes on level road (μ=0.5) from 20 m/s. Minimum stopping distance (g=10 m/s²) is:

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1. **Deceleration ($a$)**: On a level road, the only force stopping the car (excluding air resistance) is friction $f = \mu N = \mu mg$. 
   According to $F = ma$, $ma = -\mu mg \implies a = -\mu g$.
   $a = -0.5 	imes 10 = -5 	ext{ m/s}^2$.
2. **Stopping Distance ($s$)**: Using $v^2 = u^2 + 2as$ where $v=0$:
   $0 = (20)^2 + 2(-5)s$
   $10s = 400 \implies s = 40 	ext{ m}$.
Note: The mass of the car cancels out; stopping distance depends only on velocity, gravity, and the coefficient of friction.

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Friction (braking)

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